pomoc

goranspax's picture

Natrijum-hidroksid i bakar (||)-nitrit reaguju prema sledecoj jednacini 2NaOH +Cu (NO3)2---->Cu(OH)2+2NaNO3 izracunati masu NaOH koja je potrebna za reakciju sa 240.0g 10% rastvora Cu (NO3)2

milicaradenkovic's picture

Da ti ne pisem reakciju

ms=mr*w=240*0.1=24 g Ca(NO3)2

n=ms/Mr=24/164=0.1463 mol

n(NaOH)=2*n=2*0.1463=0.2926 mol

m(NaOH)=n(NaOH)*Mr=40*0.2926=11.704 g

 

JOVAN's picture

ja msm da ovako ide

2NaOH +Cu (NO3)2---->Cu(OH)2+2NaNO3

m(NaOH)=?

-----------------------------------

m(Cu(NO3)2)=240g

m%= 10%

m%= ms/mu x 100

ms= m% x mu /100

ms= 240g x 10% / 100%

ms 24g

2NaOH +Cu (NO3)2---->Cu(OH)2+2NaNO3

 

2molNaOH-------------------1mol Cu(NO3)2                               M(NaOH)= 40g/mol x2 = 80 g/mol

Xg--------------------------------24g                                                M(Cu(NO3)2) = 187g

-------------------------------------------

80g--------------------------------------187g

Xg---------------------------------------24g

187g x Xg = 80g x 24g

Xg= 80g x 24g / 187g

X=10.26gNaOH

 

milicaradenkovic's picture

Jeste ja sam umesto Cu uzela Ca .